A manufacturer ships toasters in cartons of 20. In each​ carton, they estimate a 10​% chance that any one of the toasters will need to be sent back for minor repairs. What is the probability that in a​ carton, there will be exactly 2 toasters that need​ repair?

0.2852

To find the probability that exactly 2 toasters in a carton will need repair, we can use the binomial probability formula:

P(X=k) = C(n,k) * p^k * (1-p)^(n-k)

Where:
P(X=k) = Probability of getting exactly k successes
n = Total number of trials (toasters in the carton)
k = Number of successes (toasters needing repair)
p = Probability of success in one trial (chance of a toaster needing repair)
(1-p) = Probability of failure in one trial (chance of a toaster NOT needing repair)
C(n,k) = Number of combinations of n items taken k at a time

Given:
n = 20 (toasters in the carton)
k = 2 (toasters needing repair)
p = 0.10 (10% chance)

Let's calculate the probability using the formula:

P(X=2) = C(20,2) * 0.10^2 * (1-0.10)^(20-2)

Calculating the combinations:
C(20,2) = 20! / (2!(20-2)!)
= 20! / (2!18!)
= (20*19) / (2*1)
= 190 / 2
= 190

Now, let's substitute the values:

P(X=2) = 190 * 0.10^2 * (1-0.10)^(20-2)
= 190 * 0.01 * 0.9^18

Calculating the result:

P(X=2) ≈ 0.28598

Therefore, the probability that in a carton of toasters, there will be exactly 2 toasters that need repair is approximately 0.28598 or 28.598%.

To find the probability that exactly 2 toasters in a carton need repair, we can use the concept of binomial probability.

Let's break down the problem into smaller steps:

Step 1: Determine the probability that a single toaster needs repair.
The manufacturer estimates a 10% chance that any one of the toasters will need to be sent back for minor repairs. So, the probability that a single toaster needs repair is 0.10 or 10%.

Step 2: Determine the probability that a single toaster does not need repair.
The probability that a single toaster does not need repair is the complement of the probability that it needs repair. Therefore, the probability that a single toaster does not need repair is 1 - 0.10 = 0.90 or 90%.

Step 3: Determine the probability that exactly 2 toasters in a carton need repair.
This can be calculated using the binomial probability formula:
P(X=k) = (nCk) * p^k * (1-p)^(n-k)

In this case, n is the total number of toasters in a carton (20), k is the number of toasters that need repair (2), and p is the probability that a single toaster needs repair (0.10).
Using the formula, we can calculate the probability as follows:

P(X=2) = (20C2) * (0.10)^2 * (0.90)^(20-2)

Now, let's calculate each part:

- (20C2) represents the number of ways to choose 2 toasters out of 20 to need repair. It can be calculated as:
(20C2) = 20! / (2!(20-2)!)
= 20! / (2! * 18!)
= (20 * 19) / (2 * 1)
= 190

- (0.10)^2 represents the probability that exactly 2 toasters out of 20 need repair.

- (0.90)^(20-2) represents the probability that the remaining 18 toasters out of 20 do not need repair.

Now, let's put it all together:

P(X=2) = 190 * (0.10)^2 * (0.90)^(20-2)
≈ 0.285

Therefore, the probability that in a carton, there will be exactly 2 toasters that need repair is approximately 0.285 or 28.5%.